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Indian Navy 10+2 B.Tech Cadet Magnetism & EMI — practice questions

Explore 6 free Magnetism & EMI questions in Indian Navy 10+2 B.Tech Cadet Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Magnetism & EMI questions and exercises with answers

Showing questions 1–6 of 6.

A $50$-turn coil has a magnetic flux changing through it at a uniform rate of $4 \times 10^{-3}\,\text{Wb/s}$. The induced EMF in the coil is:Read the answer and worked solution →An ideal LC circuit oscillates freely. If $L = 4\,\text{mH}$ and $C = 25\,\mu\text{F}$, the angular frequency of oscillation is:Read the answer and worked solution →A pendulum of mass m swings to amplitude angle θ. At the lowest point of swing, the TENSION in the string equals:Read the answer and worked solution →A car of mass 1000 kg moving at 36 km/h is stopped by brakes in 5 seconds. Average force exerted by brakes:Read the answer and worked solution →A body in equilibrium under several forces hasRead the answer and worked solution →A 0.5 kg ball hits a wall at 10 m/s and rebounds at 10 m/s. The impulse on the ball isRead the answer and worked solution →

Worked practice questions

A $50$-turn coil has a magnetic flux changing through it at a uniform rate of $4 \times 10^{-3}\,\text{Wb/s}$. The induced EMF in the coil is:

  1. $1.00\,\text{V}$
  2. $0.20\,\text{V}$
  3. $0.05\,\text{V}$
  4. $0.10\,\text{V}$
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Correct answer: B. $0.20\,\text{V}$

Faraday's law: $|\varepsilon| = N \dfrac{d\Phi}{dt} = 50 \times 4 \times 10^{-3} = 0.20\,\text{V}$.

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An ideal LC circuit oscillates freely. If $L = 4\,\text{mH}$ and $C = 25\,\mu\text{F}$, the angular frequency of oscillation is:

  1. $10^3\,\text{rad/s}$
  2. $10^5\,\text{rad/s}$
  3. $10^2\,\text{rad/s}$
  4. $10^4\,\text{rad/s}$
Show answer and solution

Correct answer: A. $10^3\,\text{rad/s}$

For a free LC oscillation $\omega = 1/\sqrt{LC}$. We have $LC = 4 \times 10^{-3} \times 25 \times 10^{-6}$. So $LC = 10^{-7}\,\text{s}^2$. so $\sqrt{LC} = 10^{-3.5}\,\text{s}$ and $\omega = 10^{3.5} \approx 3162\,\text{rad/s}$. Rounded to the nearest power of ten this is $10^3\,\text{rad/s}$.

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A pendulum of mass m swings to amplitude angle θ. At the lowest point of swing, the TENSION in the string equals:

  1. mg
  2. mg(2 − cos θ)
  3. mg(3 − 2cos θ)
  4. 2mg sin θ
Show answer and solution

Correct answer: C. mg(3 − 2cos θ)

**At the lowest point:** T − mg = mv²/L (centripetal). v² at bottom = 2gL(1 − cos θ) (energy conservation from amplitude). We have So T = mg + 2mg(1 − cos θ). So So T = mg(3 − 2cos θ). For θ = 60°: T = mg(3 − 1) = 2mg.

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